How do you graph #f(x) = (x-2)^2 - 1# on a coordinate graph?

Graph:

#f(x)=(x-2)^2-1# is a quadratic equation in vertex form:

#f(x)=a(x-h)^2+k#,

where:

#a=0#, #h=2#, #k=-1#

In order to graph a parabola, the vertex, x- and y-intercepts, and additional points need to be determined.

Vertex: maximum or minimum point of a parabola

The vertex is the point #(h,k)#: #(2,-1)#

Y-intercept: value of #y# when #x=0#

Substitute #y# for #f(x)# and substitute #0# for #x#. Solve for #y#.

#y=(0-2)^2-1#

#y=(-2)^2-1#

#y=4-1#

#y=3#

The y-intercept is #(0,3)#.

X-intercept: values of #x# when #y=0#

Substitute #0# for #y# and solve for #x#.

#0=(x-2)^2-1#

Expand.

#0=x^2-4x+4-1#

Simplify.

#0=x^2-4x+3#

Factor.

#0=(x-1)(x-3)#

Set each binomial equal to #0# and solve for #x#.

#x-1=0#

#x=1#

#x-3=0#

#x=3#

x-intercepts (roots): #(1,0)# and #(3,0)#

Additional point: choose value of #x#, plug it into the equation and solve for #y#.

#x=4#

Substitute #4# for #x# and solve for #y#.

#y=(4-2)^2-1#

#y=2^2-1#

#y=4-1#

#y=3#

Point: #(4,3)#

Summary

Vertex: #(2,-1)#

Y-intercept: #(0,3)#

X-intercepts: #(1,0)#, #(3,0)#

Additional point: #(4,3)#

Plot the points and sketch a parabola through them. Do not connect the dots.

graph{y=(x-2)^2-1 [-10, 10, -5, 5]}